2021,10,18 题解报告
写在前面
\(T1\) 没想出来，卒
T1
招待（entertain）
solution
对 \(W\) 进行三进制拆分，每一位是一个砝码。
如果第 \(i\) 位是 \(2\) 就将其进位（在该位置放一个物品），因为每个物品只有一个。
最后得到的一个 \(01\) 串就是放物品的最终状态。
code
/*
work by: Ariel_
Knowledge:
Time:
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define int long long
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
int stc[200], sc, a[200], b[200], W, x;
signed main(){
b[1] = 1;
for (int i = 2; i <= 35; i++) b[i] = b[i - 1] * 3;
W = read();
x = W;
while(x) {stc[++sc] = x % 3, x /= 3;}
for (int i = 1; i <= sc; i++) {
stc[i + 1] += stc[i] / 3;
stc[i] %= 3;
if(stc[i] == 2) {
stc[i + 1]++, stc[i] = 0;
a[i] = true;
}
if(stc[i + 1] > 0) sc = max(i + 1, sc);
}
for (int i = 1; i <= sc; i++) if(stc[i] == 1) cout<<b[i]<<" ";
puts("");
cout<<W<<" ";
for (int i = 1; i <= sc; i++) if(a[i]) cout<<b[i]<<" ";
puts("");
return 0;
}
T2
novel
solution
数据范围不大，分层图直接碾过去了 = =。
正解是二分路径的最大值，然后 \(bfs\) 判断是否合法。
code
/*
work by: Ariel_
Knowledge:
Time:
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXM = 2e5 + 5;
int read() {
int x = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
int n, m, k, fa[MAXN], Ans = 0x3f3f3f3f;
struct edge{int v, nxt, w;}e[MAXM << 1];
int head[MAXN], E;
void add_edge(int u, int v, int w) {
e[++E] = (edge) {v, head[u], w};
head[u] = E;
}
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
queue<int> q;
int dis[MAXN];
void work(int s) {
memset(dis, 0x3f, sizeof dis);
dis[s] = 0, q.push(s);
while(!q.empty()) {
int u = q.front(); q.pop();
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if(dis[v] > max(dis[u], e[i].w)) {
dis[v] = max(dis[u], e[i].w);
if(v % n != 0) q.push(v);
}
}
}
}
int main(){
freopen("novel.in", "r", stdin);
freopen("novel.out", "w", stdout);
n = read(), m = read(), k = read();
for (int i = 1; i <= n; i++) fa[i] = i;
for (int i = 1; i <= m; i++) {
int u = read(), v = read(), w = read();
add_edge(u, v, w), add_edge(v, u, w);
for (int j = 1; j <= k; j++){
add_edge(u + (j - 1) * n, v + j * n, 0);
add_edge(v + (j - 1) * n, u + j * n, 0);
add_edge(u + j * n, v + j * n, w);
add_edge(v + j * n, u + j * n, w);
}
if(find(u) != find(v)) fa[find(u)] = find(v);
}
if(find(1) != find(n)) {puts("-1"); return 0;}
work(1);
for(int i = 0; i <= k; i++) Ans = min(Ans, dis[n + n * k]);
cout<<Ans;
return 0;
}
chen_怡's code
二分
code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<map>
#define int long long
using namespace std;
const int N = 1e3 + 9;
const int M = 1e4 + 9;
struct node{
int last;
int to;
int dis;
}e[M<<1];
int n , m , k;
int dis[N];
int head[N] , cnt;
bool vis[N];
int l = 0 , r = 0;
int ans = -1;
int read()
{
int f = 1 , x = 0;
char s = getchar();
while(s<'0' || s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){x=(x<<1)+(x<<3)+(s^'0');s=getchar(); }
return f*x;
}
void add(int from,int to,int dis)
{
e[++cnt].last = head[from];
e[cnt].to = to;
e[cnt].dis = dis;
head[from] = cnt;
}
bool check(int x)
{
queue<int> q;
memset(dis,0x3f3f3f3f,sizeof(dis));
memset(vis,false,sizeof(vis));
dis[1] = 0;
q.push(1);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u] ; i ; i = e[i].last)
{
int v = e[i].to;
int w = e[i].dis;
if(w <= x)
{
if(dis[v] > dis[u])
{
dis[v] = dis[u];
if(!vis[v])
{
q.push(v);
vis[v] = true;
}
}
}
else if(w > x and dis[u] < k)
{
if(dis[v] > dis[u] + 1)
{
dis[v] = dis[u] + 1;
if(!vis[v])
{
q.push(v);
vis[v] = true;
}
}
}
}
}
return dis[n] <= k;
}
signed main()
{
n = read(); m = read();
k = read();
for(int i = 1 ; i <= m ; i ++)
{
int u = read();
int v = read();
int w = read();
add(u,v,w);
add(v,u,w);
r = max(r , w);
}
while(l <= r)
{
int mid = (l + r) >>1;
if(check(mid))
{
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%lld\n",ans);
return 0;
}
T3
solution
在 \([−100,100]\) 中二分得到 \(mid\)，让所有红叶的年轻程度加上该 \(mid\) 值，跑 \(Kruskal\)，直到得到最终结果;
发现修改权值之后会有重复的部分。
这个可以直接在排序的时候把红色的边放到前面就好了。
code
/*
work by: Ariel_
Knowledge:
Time:
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define int long long
using namespace std;
const int MAXN = 500005;
const int MAXM = 100005;
int read() {
int x = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
struct edge{
int w, cw, u, v, col;
bool operator < (const edge &rhs) const {
if(cw == rhs.cw) return col < rhs.col;
return cw < rhs.cw;
}
}e[MAXM];
int fa[MAXN], n, m, need;
int find(int x) {return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);}
int kruskal(int x) {
int cnt = 0, ret = 0;
for(int i = 1; i <= n; i++) fa[i] = i;
for(int i = 1; i <= m; i++){
if(!e[i].col) e[i].cw = e[i].w + x;
else e[i].cw = e[i].w;
}
sort(e + 1, e + m + 1);
for(int i = 1; i <= m; i++){
if(find(e[i].u) != find(e[i].v)){
fa[find(e[i].u)] = find(e[i].v);
if(!e[i].col) cnt++;
ret += e[i].cw;
}
}
return cnt >= need ? ret : -1;
}
int work(int l, int r) {
if(l == r) return l;
int mid = (l + r + 1) >> 1;
int x = kruskal(mid);
if(x == -1) return work(l, mid - 1);
return work(mid, r);
}
signed main() {
n = read(), m = read(), need = read();
for(int i = 1; i <= m; i++){
e[i].u = read() + 1, e[i].v = read() + 1;
e[i].w = read(), e[i].col = read();
}
int k = work(-500, 500);
printf("%lld\n", kruskal(k) - need * k);
return 0;
}
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